Q116 / 150 · MHT CET 2020 (20 Oct Shift 1) · Physics

$\quad$ Let $\overrightarrow{\mathrm{P}}=\hat{\mathrm{I}} \mathrm{P} \sin \theta-\hat{\mathrm{P}} \cos \theta$, be any vector. Another vector $\overrightarrow{\mathrm{Q}}$ which is perpendicular to $\overrightarrow{\mathrm{P}}$ is

  • A
    $(\hat{\mathrm{I}} \mathrm{Q} \sin \theta+\hat{\mathrm{j}} \mathrm{Q} \cos \theta)$
  • Correct answer
    B
    $(\hat{\mathrm{I}} \mathrm{Q} \cos \theta+\hat{\mathrm{j}} \mathrm{Q} \sin \theta)$
  • C
    $(\hat{\mathrm{I}} \mathrm{Q} \cos \theta-\hat{\mathrm{j}} \mathrm{Q} \sin \theta)$
  • D
    $(\hat{\mathrm{l}} \mathrm{P} \sin \theta+\hat{\mathrm{j}} \mathrm{P} \cos \theta)$

Solution

$\vec{P}=P \sin \theta \hat{i}-P \cos \theta \hat{j}$ also, $\vec{Q}=Q \cos \theta\hat{i}+Q \sin \theta\hat{j}$ $\therefore \vec{P} .\vec{Q}=(P \sin \theta \hat{i}-P \cos \theta \hat{j}) \cdot(Q \cos \theta \hat{i}+Qsin\theta \hat{j})$ $=P Q \sin \theta \cos \theta-P \cos \theta- Q\sin \theta$ $=0$

Hence, the correct answer is option: 'B'.

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